3.1.3 \(\int x \text {csch}^{-1}(a+b x) \, dx\) [3]

Optimal. Leaf size=75 \[ \frac {(a+b x) \sqrt {1+\frac {1}{(a+b x)^2}}}{2 b^2}-\frac {a^2 \text {csch}^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {csch}^{-1}(a+b x)-\frac {a \tanh ^{-1}\left (\sqrt {1+\frac {1}{(a+b x)^2}}\right )}{b^2} \]

[Out]

-1/2*a^2*arccsch(b*x+a)/b^2+1/2*x^2*arccsch(b*x+a)-a*arctanh((1+1/(b*x+a)^2)^(1/2))/b^2+1/2*(b*x+a)*(1+1/(b*x+
a)^2)^(1/2)/b^2

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6457, 5577, 3858, 3855, 3852, 8} \begin {gather*} -\frac {a^2 \text {csch}^{-1}(a+b x)}{2 b^2}+\frac {(a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}}{2 b^2}-\frac {a \tanh ^{-1}\left (\sqrt {\frac {1}{(a+b x)^2}+1}\right )}{b^2}+\frac {1}{2} x^2 \text {csch}^{-1}(a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcCsch[a + b*x],x]

[Out]

((a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(2*b^2) - (a^2*ArcCsch[a + b*x])/(2*b^2) + (x^2*ArcCsch[a + b*x])/2 - (a*
ArcTanh[Sqrt[1 + (a + b*x)^(-2)]])/b^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3858

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

Rule 5577

Int[Coth[(c_.) + (d_.)*(x_)]*Csch[(c_.) + (d_.)*(x_)]*(Csch[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.)*((e_.) + (
f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(e + f*x)^m)*((a + b*Csch[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Dist[f*
(m/(b*d*(n + 1))), Int[(e + f*x)^(m - 1)*(a + b*Csch[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n},
x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 6457

Int[((a_.) + ArcCsch[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[-(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Csch[x]*Coth[x]*(d*e - c*f + f*Csch[x])^m, x], x, ArcCsch[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int x \text {csch}^{-1}(a+b x) \, dx &=-\frac {\text {Subst}\left (\int x \coth (x) \text {csch}(x) (-a+\text {csch}(x)) \, dx,x,\text {csch}^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \text {csch}^{-1}(a+b x)-\frac {\text {Subst}\left (\int (-a+\text {csch}(x))^2 \, dx,x,\text {csch}^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a^2 \text {csch}^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {csch}^{-1}(a+b x)-\frac {\text {Subst}\left (\int \text {csch}^2(x) \, dx,x,\text {csch}^{-1}(a+b x)\right )}{2 b^2}+\frac {a \text {Subst}\left (\int \text {csch}(x) \, dx,x,\text {csch}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a^2 \text {csch}^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {csch}^{-1}(a+b x)-\frac {a \tanh ^{-1}\left (\sqrt {1+\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {i \text {Subst}\left (\int 1 \, dx,x,-i (a+b x) \sqrt {1+\frac {1}{(a+b x)^2}}\right )}{2 b^2}\\ &=\frac {(a+b x) \sqrt {1+\frac {1}{(a+b x)^2}}}{2 b^2}-\frac {a^2 \text {csch}^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {csch}^{-1}(a+b x)-\frac {a \tanh ^{-1}\left (\sqrt {1+\frac {1}{(a+b x)^2}}\right )}{b^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 110, normalized size = 1.47 \begin {gather*} \frac {(a+b x) \sqrt {\frac {1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}+b^2 x^2 \text {csch}^{-1}(a+b x)-a^2 \sinh ^{-1}\left (\frac {1}{a+b x}\right )-2 a \log \left ((a+b x) \left (1+\sqrt {\frac {1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}\right )\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCsch[a + b*x],x]

[Out]

((a + b*x)*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2] + b^2*x^2*ArcCsch[a + b*x] - a^2*ArcSinh[(a + b*x)^
(-1)] - 2*a*Log[(a + b*x)*(1 + Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])])/(2*b^2)

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Maple [A]
time = 0.22, size = 97, normalized size = 1.29

method	result	size
derivativedivides	\(\frac {-\mathrm {arccsch}\left (b x +a \right ) a \left (b x +a \right )+\frac {\mathrm {arccsch}\left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\frac {\sqrt {\left (b x +a \right )^{2}+1}\, \left (2 a \arcsinh \left (b x +a \right )-\sqrt {\left (b x +a \right )^{2}+1}\right )}{2 \left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}+1}{\left (b x +a \right )^{2}}}}}{b^{2}}\)	\(97\)
default	\(\frac {-\mathrm {arccsch}\left (b x +a \right ) a \left (b x +a \right )+\frac {\mathrm {arccsch}\left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\frac {\sqrt {\left (b x +a \right )^{2}+1}\, \left (2 a \arcsinh \left (b x +a \right )-\sqrt {\left (b x +a \right )^{2}+1}\right )}{2 \left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}+1}{\left (b x +a \right )^{2}}}}}{b^{2}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccsch(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b^2*(-arccsch(b*x+a)*a*(b*x+a)+1/2*arccsch(b*x+a)*(b*x+a)^2-1/2*((b*x+a)^2+1)^(1/2)*(2*a*arcsinh(b*x+a)-((b*
x+a)^2+1)^(1/2))/(b*x+a)/(((b*x+a)^2+1)/(b*x+a)^2)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccsch(b*x+a),x, algorithm="maxima")

[Out]

1/2*I*a*(log(I*(b^2*x + a*b)/b + 1) - log(-I*(b^2*x + a*b)/b + 1))/b^2 + 1/4*(2*b^2*x^2*log(sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 1) + 1) - (a^2 - 1)*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*(b^2*x^2 - a^2)*log(b*x + a))/b^2 + inte
grate(1/2*(b^2*x^3 + a*b*x^2)/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 1), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (65) = 130\).
time = 0.38, size = 285, normalized size = 3.80 \begin {gather*} \frac {b^{2} x^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) - a^{2} \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a + 1\right ) + a^{2} \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a - 1\right ) + 2 \, a \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right ) + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccsch(b*x+a),x, algorithm="fricas")

[Out]

1/2*(b^2*x^2*log(((b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/(b*x + a)) - a^
2*log(-b*x + (b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a + 1) + a^2*log(-b*x +
(b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a - 1) + 2*a*log(-b*x + (b*x + a)*s
qrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a) + (b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 +
1)/(b^2*x^2 + 2*a*b*x + a^2)))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {acsch}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acsch(b*x+a),x)

[Out]

Integral(x*acsch(a + b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccsch(b*x+a),x, algorithm="giac")

[Out]

integrate(x*arccsch(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {asinh}\left (\frac {1}{a+b\,x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asinh(1/(a + b*x)),x)

[Out]

int(x*asinh(1/(a + b*x)), x)

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